
Re: Quasifields
Posted:
Jul 18, 2011 8:13 AM


On Mon, 18 Jul 2011, William Elliot wrote: > On Mon, 18 Jul 2011, David Hartley wrote: >>>>>>> >>>>>>>> In Dauben's biography of Abraham Robinson (page 103 of the >>>>>>>> first printing) a quasifield is said to be a commutative >>>>>>>> field with distribution >>>>>>>> >>>>>>>> a(b + c) = ab + ac (1) >>>>>>>> >>>>>>>> replaced by >>>>>>>> a(b_1 + b_2 + ... + b_n) = ab_1 + ab_2 + ... + ab_n. (2) >>> >>>>> Thus, in a quasifield, the law x*0 = 0 holds iff the >>>>> distributive law holds. >>>>> >>> Define as usual for n in N, na = sum(j=1,n) a. >>> >>> a0 = a * n0 = n(a0); (n1)(a0) = 0 >>> >>> Thus quasi distributivity (QD). >>> a(b + c) = a(b + c + (n2)0) = ab + ac + (n2)(a0) = ab + ac  a0 >>> >>> Does (1)a = a hold in a quasi field? >> >> It requires proof. It does hold for n=3: >> >> a*b = a*(b + b + b) >> = a*b + a*b + a*(b) >> >> so a*b + a*(b) = 0, a*(b) = (a*b) >> Using QD a0 = a(1  1) = a1 + a(1)  a0 a(1) = a0 + a0  a
If n = 3: a0 = a(0 + 0 + 0) = a0 + a0 + a0 a0 + a0 = 0 a(1) = a
> In general for n = 2k + 1: > > a0 = a(kb + k(b) + 0) = k(ab) + k(a(b) + a0 > > k(a(b)) = k(ab) > >> But that's no help for n even. > > For n = 2k: > a0 = a(kb + k(b)) = k(ab) + k(a(b)) > > k(a(b)) = a0  k(ab) > > Maybe a0 is an infinitesimal. > >>> Assume it does. >>> a0 = a(1  1) = a1 + a(1)  a0 = a  a  a0 >>> a0 + a0 = 0 >>> a0 = a * n0 = n(a0) >>> >>> If n is even, then a0 = 0 and we've a field.
Misc. (1)0 = 0*0 + 0*0 (1 + 1)a = a1 + a1  a0 = a + a  a0 (1 + 1)0 = (0*0)
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